我需要找到多边形内和多边形上的所有晶格点。
输入:
from shapely.geometry import Polygon, mapping
sh_polygon = Polygon(((0,0), (2,0), (2,2), (0,2)))
输出:
(0, 0), (1, 0), (2, 0), (0, 1), (1, 1), (2, 1), (0, 2), (1, 2), (2, 2)
[请提出是否有办法使用或不使用Shapely来获得预期的结果。
我已经编写了这段代码,该代码给出了多边形内的点,但没有给出点。还有一种更好的方法可以执行相同的操作:
from shapely.geometry import Polygon, Point
def get_random_point_in_polygon(poly):
(minx, miny, maxx, maxy) = poly.bounds
minx = int(minx)
miny = int(miny)
maxx = int(maxx)
maxy = int(maxy)
print("poly.bounds:", poly.bounds)
a = []
for x in range(minx, maxx+1):
for y in range(miny, maxy+1):
p = Point(x, y)
if poly.contains(p):
a.append([x, y])
return a
p = Polygon([(0,0), (2,0), (2,2), (0,2)])
point_in_poly = get_random_point_in_polygon(p)
print(len(point_in_poly))
print(point_in_poly)
输出:
poly.bounds: (0.0, 0.0, 2.0, 2.0)
1
[[1, 1]]
我简化了我的问题。实际上,我需要找到带有角的正方形内和上的所有点:(77,97),(141,101),(136,165),(73,160)。
是否没有一个函数可以找到位于[
p1 = (x1, y1)
p2 = (x2, y2)
xdiff = x2 - x1
ydiff = y2 - y1
# Find the line's equation, y = mx + b
m = ydiff / xdiff
b = y1 - m*x1
for xval in range(x1+1, x2):
yval = m * xval + b
if int(yval) == yval:
# add (xval, yval) to your list of points
我已将细节留给您:请确保x1